Question

A substance with specific heat 0.2 cal/(g ⋅°C) which is initially at 70 °C comes into contact with the same amount of another specific heat substance 0.5 cal/(g ⋅°C). The final temperature when equilibrium is reached is 46 °C. Calculates the initial temperature of the second substance.

Answer 1

Explanation:

Heat energy given by metal piece =m⋅c⋅ΔT

1

=20×0.3×(100−22)

=468J

Heat energy gained by water =m

w

×c

w

×ΔT

2

=m

w

×4.2×(22−20)

=m

w

×8.4J

Heat energy gained by calorimeter =m

c

×c

c

×ΔT

2

=50×0.42×(22−20)

=42J

By principle of calorimeter:

Heat lost = Heat gained

Heat energy given by metal = Heat energy gained by water + Heat energy gained by calorimeter

468=m

w

×8.4+42

m

w

=

8.4

468−42

=50.7g

Therefore, the mass f water used in the calorimeter is 50.7g.

Answer 2

Explanation:

We now introduce two concepts useful in describing heat flow and temperature change. The heat capacity (C) of a body of matter is the quantity of heat (q) it absorbs or releases when it experiences a temperature change (ΔT) of 1 degree Celsius (or equivalently, 1 kelvin)

C=

q

ΔT

Heat capacity is determined by both the type and amount of substance that absorbs or releases heat. It is therefore an extensive property—its value is proportional to the amount of the substance.

For example, consider th